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Find the sum of the following geometric progressions:

(i) 2, 6, 18, … to 7 terms

(ii) 1, 3, 9, 27, … to 8 terms

(iii) 1, -1/2, 1/4, -1/8, …

(iv) (a2 – b2), (a – b), (a - b)/(a + b), … to n terms

(v) 4, 2, 1, 1/2 … to 10 terms

1 Answer

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Best answer

(i) 2, 6, 18, … to 7 terms

We know that, sum of GP for n terms = a(rn – 1)/(r – 1)

Given:

a = 2, r = t2/t1 = 6/2 = 3, n = 7

Now let us substitute the values in

a(rn – 1)/(r – 1) = 2 (37 – 1)/(3 - 1)

= 2 (37 – 1)/2

= 37 – 1

= 2187 – 1

= 2186

(ii) 1, 3, 9, 27, … to 8 terms

We know that, sum of GP for n terms = a(rn – 1)/(r – 1)

Given:

a = 1, r = t2/t1 = 3/1 = 3, n = 8

Now let us substitute the values in

a(rn – 1)/(r – 1) = 1 (38 – 1)/(3 - 1)

= (38 – 1)/2

= (6561 – 1)/2

= 6560/2

= 3280

(iii) 1, -1/2, ¼, -1/8, …

We know that, sum of GP for infinity = a/(1 – r)

Given:

a = 1, r = t2/t1 = (-1/2)/1 = -1/2

Now let us substitute the values in

a/(1 – r) = 1/(1 – (-1/2))

= 1/(1 + 1/2)

= 1/((2 + 1)/2)

= 1/(3/2)

= 2/3

(iv) (a2 – b2), (a – b), (a - b)/(a + b), … to n terms

We know that, sum of GP for n terms = a(rn – 1)/(r – 1)

Given:

a = (a2 – b2), r = t2/t1 = (a - b)/(a2 – b2) = (a - b)/(a - b) (a + b) = 1/(a + b), n = n

Now let us substitute the values in

a(rn – 1)/(r – 1) =

(v) 4, 2, 1, 1/2 … to 10 terms

We know that, sum of GP for n terms = a(rn – 1)/(r – 1)

Given:

a = 4, r = t2/t1 = 2/4 = 1/2, n = 10

Now let us substitute the values in

a(rn – 1)/(r – 1) = 4 ((1/2)10 – 1)/((1/2) - 1)

= 4 ((1/2)10 – 1)/((1 - 2)/2)

= 4 ((1/2)10 – 1)/(-1/2)

= 4 ((1/2)10 – 1) × -2/1

= -8 [1/1024 - 1]

= -8 [1 – 1024]/1024

= -8 [-1023]/1024

= 1023/128

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